The above congruence has a solution if and only if its discrimant is a quadradic residue.
Since .
Apply Euler's Criterion to determine if something is a quadradic residue we get that,
Thus, is never divisible by 5, thus how can it be divisible by ?
Thus, we have shown it is impossible.
Q.E.D.

To remind you Euler's Criterion states, is a quadradic residue of if and only if,